# If the work required to stretch a spring 1 foot beyond its natural length is 12 foot-pounds, how much work is needed to stretch it 9 inches beyond its natural length?

Aug 21, 2014

The answer is $\frac{27}{4}$ ft-lbs.

Let's look at the integral for work (for springs):

$W = {\int}_{a}^{b} k x \setminus \mathrm{dx} = k \setminus {\int}_{a}^{b} x \setminus \mathrm{dx}$

Here's what we know:

$W = 12$
$a = 0$
$b = 1$

So, let's substitute these in:

$12 = k {\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}$
$12 = k \left(\frac{1}{2} - 0\right)$
$24 = k$

Now:

9 inches = 3/4 foot = $b$

So, let's substitute again with $k$:

$W = {\int}_{0}^{\frac{3}{4}} 24 x \mathrm{dx}$
$= \frac{24 {x}^{2}}{2} {|}_{0}^{\frac{3}{4}}$
$= 12 {\left(\frac{3}{4}\right)}^{2}$
$= \frac{27}{4}$ ft-lbs

Always set up the problem with what you know, in this case, the integral formula for work and springs. Generally, you will need to solve for $k$, that's why $2$ different lengths are provided. In the case where you are given a single length, you're probably just asked to solve for $k$.

If you are given a problem in metric, be careful if you are given mass to stretch or compress the spring vertically because mass is not force. You will have to multiply by 9.8 $m {s}^{- 2}$ to compute the force (in newtons).